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Proof step 4 from proof put-in (previous step)

∀ a0, d0, a, d. (a0 × d0 ∈ ax0[a, d] ↔ a0 × d0 = a × d) ∨ a0 × d0 ∈ ax0 ⊦ a0 × d0 ∈ (p + ax0)[a, d] ↔ a0 × d0 = a × d, a0 × d0 ∈ p + ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a1, d1, a2, d2. (a1 × d1 ∈ ax0[a2, d2] ↔ a1 = a2 ∧ d1 = d2) ∨ a1 × d1 ∈ ax0 ⊦ a0 × d0 ∈ (p ' + ax0)[a, d] ↔ a0 × d0 = a × d, a0 × d0 = p, a0 × d0 ∈ ax0

Rule argument

Additional information

Used simplifier rules: ⊦ a × d = a0 × d0 ↔ a = a0 ∧ d = d0;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ ¬ p ∈ p0 ' + ax ↔ p ≠ p0 ∧ ¬ p ∈ ax;% ⊦ p + ax = p ' + ax;